-4(2y)+11y=15 how do I solve this equation

Answers

Answer 1
Answer:

Answer:

y=5

Step-by-step explanation:

multiply the -4 with 2y

combine the -8y with the positive 11y getting 3y

divide the 3 to get y by itself

y=5


Related Questions

The sum of the values of five coins is $0.85. What are the five coins?PLEASE HELP ME!!!!!!
Fraction 33/100 as decimal
Donald is planting two types of saplings in his garden.sapling 1 is4 feet tall and grows at the rate of 16 inches per year. Sapling 2 is 5 feet tall and grows at the rate of 12 inches per year.after how many yeas will these saplings be of the same height?
Find the area of the shaded region. Round your answer to the nearest tenth.
Which is the equation of a line that has a slope of 1/2 and passes through point (2, -3)?

I need to simplify the circled questions and explain how I got my answers.

Answers

7) 2^3 / 2(3+5)
8/2(5)
4/5
.8

10) 8(4-1) x 4
8(3) x 4
24x4
96

12) 9^2 (1+1) + 7
81 (2) + 7
162+7
169

14) 8^2 -1 +4 /2
64-1+4/2
67/2
33.5

18) 11^2 + 2^3 -3 x 4
121 + 8 - 3 x 4
129 -12
117

#72^3÷2(3+2)2^2×(3+2)4×520

Carlos bought 1 pound of carrots how many ounces of carrots did he buy

Answers

There are 16 ounces per lb and since he only has 1 lb, then
16 × 1 = 16 ounces
16 OZ, since one pound equals 16 ounces

PLEASEEE HELPP The water usage at a car wash is modeled by the equation W(x) = 3x3 + 4x2 − 18x + 4, where W is the amount of water in cubic feet and x is the number of hours the car wash is open. The owners of the car wash want to cut back their water usage during a drought and decide to close the car wash early two days a week. The amount of decrease in water used is modeled by D(x) = x3 + 2x2 + 15, where D is the amount of water in cubic feet and x is time in hours. Write a function, C(x), to model the water used by the car wash on a shorter day. C(x) = 2x3 + 2x2 − 18x − 11
C(x) = 3x3 + 2x2 − 18x + 11
C(x) = 3x3 + 2x2 − 18x − 11
C(x) = 2x3 + 2x2 − 18x + 11

Answers

We have been given that the water usage at a car wash is modeled by the equation W(x) = 3x^3+4x^2-18x + 4, where W is the amount of water in cubic feet and x is the number of hours the car wash is open.  

The amount of decrease in water used is modeled by D(x) = x^3 + 2x^2 + 15, where D is the amount of water in cubic feet and x is time in hours.

The function C(x) will be difference of W(x) and D(x) that is C(x)=W(x)-D(x).

Upon substituting both function values in above formula, we will get:

C(x)=3x^3+4x^2-18x + 4-(x^3 + 2x^2 + 15)

Let us remove parenthesis.

C(x)=3x^3+4x^2-18x + 4-x^3-2x^2-15

Combine like terms.

C(x)=3x^3-x^3+4x^2-2x^2-18x+4-15

C(x)=2x^3+2x^2-18x-11

Therefore, the function C(x)=2x^3+2x^2-18x-11 represents the water used by the car wash on a shorter day and option A is the correct choice.

Answer:

a

Step-by-step explanation:

the Formula A = LW gives the area A of a rectangle with length L and width W. What is the area in square feet of a United States flag with a length of 12 feet and a width of 8 feet?

Answers

A = L x W
A = 12 x 8
A = 96 square feet
Your answer will be 96 feet

The area of a rectangle is 4.9 square units. The length is 2.5 units and the width is y units. Which statements are true? Select each correct answer. A. 2.5y=4.9 B. y=1.96 C. 4.9x=2.5 D. y=12.25

Answers

The answer is A because the area is all the points added together so you need to times it by Y and you get the area which is 4.9
A and B. A is xy=A (4.9) : 2.5*y=4.9 and B is A/x=y so 4,9/2,5=1,96 which is correct

The paths of two runners cross at a stop sign. One runner is heading south at a constant rate of 6.5 miles per hour while the other runner is heading west at a constant rate of 7 miles per hour. How fast is the distance between them changing after 10 minutes?

Answers

Answer:

The distance between them changing after 10 minutes will be 9.553 mph.

Step-by-step explanation:

The paths of two runners cross at a stop sign (O). One runner is heading south at a constant rate of 6.5 miles per hour towards A while the other runner is heading west at a constant rate of 7 miles per hour towards B.

So, after 10 minutes the first runner covers a distance of 6.5 * (10)/(60) = 1.083 miles and the second runner covers a distance of 7 * (10)/(60) = 1.167 miles.

Therefore, after 10 minutes their distance will be \sqrt{(1.083)^(2) + (1.167)^(2)} = 1.592 miles.

Now, the distance between them is given by

AB² = OA² + OB²

Now, differentiating this equation with respect to time t (in hours) we get

2(AB) (d(AB))/(dt) = 2(OA) (d(OA))/(dt) + 2(OB) (d(OB))/(dt)

(AB) (d(AB))/(dt) = (OA) (d(OA))/(dt) + (OB) (d(OB))/(dt)

1.592 * (d(AB))/(dt) = 1.083 * 6.5 +  1.167 * 7 = 15.208

(d(AB))/(dt) = 9.553 mph.

Therefore, the distance between them changing after 10 minutes will be 9.553 mph. (Answer)

Final answer:

The distance between the two runners is not changing after 10 minutes.

Explanation:

To find the rate of change of the distance between the two runners, we can use the concept of relative velocity. The distance is changing due to the motion of both runners, so we need to find the rate at which each runner is approaching or moving away from the other. Since one runner is heading south and the other is heading west, their velocities are perpendicular to each other. We can use the Pythagorean theorem to find their combined velocity and then calculate the rate of change of the distance between them.

Let's consider the southward runner as Runner A and the westward runner as Runner B. The velocity of A is 6.5 miles per hour, and the velocity of B is 7 miles per hour. After 10 minutes, the distance traveled by A can be calculated as (6.5 miles/hour) * (10/60) hours = 1.083 miles. The distance traveled by B can be calculated as (7 miles/hour) * (10/60) hours = 1.167 miles.

Using the Pythagorean theorem, we can calculate the distance between the two runners after 10 minutes:

Distance = sqrt((1.083 miles)^2 + (1.167 miles)^2) ≈ 1.563 miles

To find the rate of change of the distance between them, we can differentiate the equation for the distance with respect to time:

d(Distance)/dt = (1/2)*((2*(1.083 miles)*(0))/(sqrt((1.083 miles)^2 + (1.167 miles)^2))) + (1/2)*((2*(1.167 miles)*(0))/(sqrt((1.083 miles)^2 + (1.167 miles)^2))) = 0

Therefore, the distance between the two runners is not changing after 10 minutes.

Learn more about Relative velocity here:

brainly.com/question/34025828

#SPJ3